FP1: Chapter 1Complex Numbers

Dr J Frost (jfrost@tiffin.kingston.sch.uk)

Last modified: 2nd January 2014

i i captain!

Here’s something that someone has almost certainly spoiled for you already…

𝑖=√−1?

Complex versus imaginary numbers

Imaginary number: of form

Complex number: of form

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3+4 𝑖Imaginary partReal part? ?

Putting complex numbers in the preferred form

We tend to ensure real and imaginary components are grouped together.(a, b and c are real constants)

a + 3i – 4 + bi = a – 4 + (3 + b)i2a – 3bi + 3 – 6ci = 2a+3 – 3(b + 2c)i

Just like we’d write 6 instead of 6, the i appears after any real constants, so we might write 5ki or i.

An exception is when we involve a function.e.g. i sin and i√3Why? This avoids ambiguity over whether the function is being applied to the i.

Convention 1

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Convention 2

In the same way that we often initially use x and y as real-valued variables, we often use z first to represent complex values, then w.

Why complex numbers?Complex numbers were originally introduced by the Italian mathematician Cardano in the 1500s to allow him to represent the roots of polynomials which weren’t ‘real’. They can also be used to represent outputs of functions for inputs not in the usual valid domain, e.g. Logs of negative numbers, or even the factorial of negative numbers!Some other major applications of Complex Numbers:

Fractals1

A Mandelbrot Set is the most popular ‘fractal’. For each possible complex number c, we see if zn+1 = zn

2 + c is not divergent (using z0 = 0), leading to the diagram on the right. Coloured diagrams can be obtained by seeing how quickly divergence occurs for each complex c (if divergent). This is called an Argand

Diagram. We’ll use it later.Analytic Number Theory2

Number Theory is the study of integers. Analytic Number Theory treats integers as reals/complex numbers to use other (‘analytic’) methods to study them. For example, the Riemann Zeta Function allows complex numbers as inputs, and is closely related to the distribution of prime numbers.

Physics and Engineering3

Used in Signal Analysis, Quantum Mechanics, Fluid Dynamics, Relativity, Control Theory...

Manipulation and application

Solve

Solve Using the quadratic formula:

Simplify

Simplify (8 + i)(3 – 2i)

= 24 – 16i + 3i – 2i2

= 24 – 16i + 3i + 2= 26 – 13i

Here’s a flavour of some of things you’ll be initially expected to do...

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MultiplyingExample: Let z1 = 1 + 3i and z2 = 5 – 2i, find z1z2.

(1 + 3i)(5 – 2i) = 5 – 2i + 5i – 6i2

= 5 – 2i + 5i + 6= 11 + 3i

Calculate z2 for the following z.

The quick way to think about this is that i2 reverses the sign.

a) z = 1 + i z2 = 2i b) z = 1 – i z2 = -2ic) z = 3 + 2i z2 = 5 + 12id) z = 7 – 4i z2 = 33 – 56ie) z = -3 + 3i z2 = -18if) z = a + bi z2 = a2 – b2 + 2abi

Calculate z1z2.a) z1 = 1 + i z2 = 1 – i z1z2 = 2b) z1 = 2 + i z2 = 2 – 2i z1z2 = 6 – 2ic) z1 = 3 + 2i z2 = 4 – 3i z1z2 = 18 – i d) z1 = -3 + i z2 = 5 + 7i z1z2 = -22 –

16ie) z1 = 7 – 3i z2 = 3 – 7i z1z2 = -58if) z1 = -1 – i z2 = 9 + 8i z1z2 = -1 –

17i2

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4 By using a Binomial Expansion, determine (1 + i)5.= 1 + 5i + 10i2 +10i3 + 5i4 + i5

= 1 + 5i – 10 – 10i + 5+ i= -4 – 4i

What is the value of:a) i100 = (i4)25 = 125 = 1b) i2003 = i2000 x i3 = 1 x i3 = -i

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Evaluate the following:a) √(-16) = 4ib) √(-25) = 5ic) √(-3) = i√3d) √(-7) = i√7e) √(-8) =2i√2

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Complex Conjugates

Remember in C1 how we rationalised the denominator?

Suppose x1 = a + √b and x2 = a - √bThen:

x1x2 = a2 – bx1 + x2 = 2a

Both these results are rational.

The same trick works with complex numbers.

! If z = x + yi, then we define z* = x – yi.z* is known as the complex conjugate of z.

z z* = x2 + y2

z + z* = 2xwhich are both real.

The fact the first is real will help us with dividing complex numbers.We’ll see the significance of the second later.

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Quickfire Questions

Given z, determine z z*.

z = 3 + 2i z z* = 13z = 3 – 2i z z* = 13z = 5 + 4i z z* = 41z = 1 + i z z* = 2z = 4 – 2i z z* = 20

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Dividing

__26__2 + 3i

(2 – 3i) ( )(2 – 3i)

= 52 – 78i13

= 4 – 6i

Click to Brosolve

Exercises

1 - i1 + i

Put all the following in the form a + bi.

_10_3 + i

_1_1 + i

8 – 4i1 – 3i

= 3 - i

= -i

Note that mark schemes permit the single fraction.

1 - i2

= 10 + 3i1 + 2i

16 – 17i5

=

_i_1 – i

-1 + i2

=

14 – 5i3 – 2i

= 4 + i

= 2 + 2i

a + ia – i

a2 – 1 + 2aia2 + 1

=

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Edexcel June 2013 (Retracted)Given that (2 + i)(z + 3i) = 10 – 5i, find z, giving your answer in the form a + bi.z = 3 – 7i (either by dividing 10 – 5i by 2 + i and subtracting 3i, or replacing z with a + bi before expanding and comparing real and imaginary parts).

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Argand Diagrams

Argand Diagrams are a well of geometrically representing complex numbers.

z = x + yiThe x-axis is the real component.The y-axis is the imaginary component.

Im[z]

Re[z]-3 -2 -1 1 2 3

3

2

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-1

-2

-3

1 + i 2 – 3i

-1 + 2i -3 – i

Click to move.

Argument and ModulusIn FP2, you’ll encounter something called ‘polar coordinates’. This is an alternative way of representing coordinate, which instead of using the (x,y) position (known as a Cartesian coordinate), uses the distance from the origin and the angle.

Im[z]

Re[z]-3 -2 -1 1 2 3

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-2

-3

z = 2 + 3i

Distance from origin:

(Don’t write anything down yet!)

√(22 + 32) = √13This is known as the modulus of z, and we write

|2 + 3i| = √13

Angle: (anticlockwise from the real axis)

Using trigonometry, we have (in radians):tan-1(3/2) = 0.983This is known as the argument of z, and we write:

arg(2 + 3i) = 0.983

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Argument and Modulus

Im[z]

Re[z]arg z

!If |z

|

Argument of z:Usual range - < arg z ≤ . Known as the principal argument if in this range.

when Use trig common sense for other

quadrants.

Modulus of z:

Check your Understanding

z |z| arg(z)

1 1 0

-i 1 -/2

-1 + i √2 3/4

-5 – 2i √29 -2.761

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Exercises

z |z| arg(z)

-1 1

i 1 /2

1 + i √2 p/4

1 + 2i √5 1.107

1 – 2i √5 -1.107

-1 + 2i √5 2.034

-1 – 2i √5 -2.034

3 + 4i 5 0.927

-5 + 12i 13 1.966

1 – i√3 √7 -/3

Give exact answers where possible, otherwise to 3dp.

z |z| arg(z)

-√3 - i √7 -5/6

3 + i √10 0.322

2 – 5i √29 -1.190

-4 + 3i 5 2.498

-1 – 4i √17 -1.816

Given that arg(3 + a + 4i) = p/3 and that a is real, determine a.4/(3 + a) = √3. Thus a = (4/√3) – 3

Give that arg(5 + i + ai) = p/4 and that a is real, determine a.(a + 1)/5 = 1. So a = 4

Given that arg(santa + 3 + 2i) = /2 and that santa is real, determine santa.santa = -3

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Argument-modulus form

What’s the point of |z| and arg(z)?We’ll see in FP2 that we can express a complex number as ., where we’ll also see the infamous . But for now, let’s get halfway there...

Im[z]

Re[z]arg z

|z|

Suppose that r = |z| and = arg(z).How could we express z in terms of r and o?

!z = x + yi = r cos + r i sin = r(cos + i sin )

This is known as the modulus-argument form (or the polar form) of z.

z |z| arg(z) Mod-arg form

-1 1 z = cos + i sin

i 1 /2 z = cos(/2) + i sin(/2)

1 + i √2 p/4 z = √2(cos(/4) + i sin(/4))

1 + 2i √5 1.107 z = √5(cos 1.107 + i sin 1.107)

1 – 2i √5 -1.107 z = √5(cos(-1.107) + i sin(-1.107))

-1 + 2i √5 2.034 z = √5(cos 2.034 + i sin 2.034)

-1 – 2i √5 -2.034 z = √5(cos(-2.034) + i sin(-2.034))

3 + 4i 5 0.927 z = 5(cos 0.927 + i sin 0.927)

-5 + 12i 13 1.966 z = 13(cos 1.966+ i sin 1.966)

1 – i√3 √7 -/3 z = √7(cos(-/3) + i sin(-/3))

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Question 1 from earlier...

Argument-modulus form

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|z1z2| and arg(z1z2)

When we multiply two complex numbers, we multiply their moduli, and we add their arguments.

So can we prove this, i.e. that: and

The following C3 identities may be useful:

Let and .

Then

So and

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Comparing coefficientsIf two complex numbers are equal, then clearly both their real and imaginary parts are equal. i.e. if then and .

Given that where and are real, find the value of and .

Expanding: So and .Solving: . We could have also found

Calculate .

So and

So

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Exercise 1G

where and are real.Find the value of and of .

where is real. Find the value of .

Find real and such that

Find the square roots of

Find the square roots of

Find the square roots of

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Roots of polynomialsAny cubic (with coefficient of x3 of 1) can be expressed as:

y = (x – α)(x – )(x – )where α, and are the roots.

However, these 3 roots may not necessarily all be real...

α α

All 3 roots are real. 1 real root, 2 complex roots.

Are there any other possibilities?No: since cubics have a range of – to +, it must cross the x axis. And it can’t cross an even number of times, otherwise the cubic would start and end in the same vertical direction.

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3 real roots (one repeated)? ?

Roots of polynomials

Classic Question:x = 2 is one of the roots of the polynomial x3 – x – 6 Find the other two roots.

Use polynomial division (as per C2) to divide x3 – x – 6 by (x – 2).This gives x2 + 2x + 3.Using the quadratic formula, we obtain the roots:

x = -1 i√2

Your Go:x = -1/2 is one of the roots of the polynomial 2x3 – 5x2 + 5x + 4.Find the other two roots.

We could divide by (x + ½), but it would be cleaner to divide by (2x + 1).This gives x^2 – 3x + 4. Using the quadratic formula, we obtain the roots:

x = - ½ (3 i√7)

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Relationship between complex roots

For the first cubic, we found the two complex roots were:x = -1 i√2

What is the relationship between these two roots?They’re complex conjugates.

Can we prove this will always be the case for cubics with real coefficients?Suppose a and b are the two complex roots. Then (x – α)(x – β) in the expansion must give real coefficients.

(x – α)(x – β) = x2 – (a + β)x + αβTherefore α + β and αβ must both be real. We saw earlier that this is satisfied when α and β are complex conjugates.(Although technically our proof isn’t complete, because we’ve shown that the roots being complex conjugates is a sufficient condition for the coefficients of the cubic being real, not a necessary one. However, we know the imaginary components of α and β must be the same but negative of each other, so that in α + β they cancel out, and we can proceed from there to show the real components of α and β must be the same).

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In summary, complex roots of polynomials always come in complex conjugate pairs.

Finding other roots

Another classic type of exam question:-1 + 2i is one of the roots of the cubic x3 – x2 – x – 15.Find the other two roots.

Other complex root is the complex conjugate: -1 – 2i.Now expand (x – (-1 + 2i))(x – (-1 – 2i))

= x2 – (-1 – 2i)x – (-1 + 2i)x + (-1 + 2i)(-1 – 2i)= x2 + 2x + 5

We can now either use polynomial division or “Factor Theorem with trial and error” to establish the real root as 3.

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Your turn:2 – i is one of the roots of the cubic x3 – 11x + 20. Find the other two roots.

2 + i is other complex root.(x – (2 + i))(x – (2 – i)) = x2 – 4x + 5.Dividing we get (x + 4), so real root is -4.

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QuarticsAny quartic (with coefficient of x4 of 1) can be expressed as:

y = (x – α)(x – )(x – )(x – δ)where α, , and δ are the roots.

α δ α

Possibility 1:4 real roots (some potentially repeated)

Possibility 2:2 real roots, pair of complex conjugate roots.

Possibility 3:No real roots. Two pairs of complex conjugate roots.

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Any other possibilities?No. On y-axis, both ends of line must either be both positive infinity or both negative. x-axis must therefore be crossed an even number of times, so an even number of real roots.?

Quartics

Edexcel Jan 2013 (Retracted)

a) 4x2 + 9 = 0x2 = -9/4, x = i√(3/2)

x2 – 2x + 5 = 0Using the quadratic formula, x = 1 2i

Im[z]

Re[z]

b)Since real roots appear on the x-axis and complex roots come in conjugate pairs, we know we’ll have symmetry in the line y = 0

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Exercise 1H

Given that 1 + 2i is one of the roots of a quadratic equation with real coefficients, find the equation.

Given that , where is real, is one of the roots of a quadratic equation with real coefficients, find the equation.

Show that is a root of the equation .Hence solve the equation completely.Roots are 3, ,

Given that is one of the roots of the equation , solve the equation completely.Roots are 4, -4 + i and -4 - i

Given that is one of the roots of the equation , solve the equation completely.Roots are 2 + 3i, 2 – 3i, -3 + i and -3 – i

Find the four roots of the equation . Show these roots on an Argand Diagram.Roots are 2, -2, 2i, -2i

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